I am confused on how to find the output resistance of transistor amplifiers in general, but more specifically when it's a combination of different resistances.

Here's an example (concrete) of the source follower (from Sedra Smith 6th ed.): Source follower and small signal equivalent

Could someone give me a purely analytical solution on how they came to the conclusion the output resistance \$ R_o \$ is \$ \frac{1}{g_m} \parallel r_o \$ in the small-signal model?

Beside, what is the general method to get the output resistance? Do we cancel \$ v_{sig} \$? Do we replace the output terminal with a voltage source \$ V_o \$ and calculate the thevenin resistance \$ \frac{V_o}{i_{o}} \$ ? I'm not sure exactly what sources we keep and what we cancel, etc.


  • \$\begingroup\$ I noticed that on this unrelated question: electronics.stackexchange.com/questions/93428/… I gave what appears to be a reasonable answer yet you have neither accepted the answer nor raised a comment to ask further questions about my answer. What should I conclude? \$\endgroup\$
    – Andy aka
    Jan 23, 2016 at 0:38
  • \$\begingroup\$ That was long ago. At the time I wrote the question I wasn't even familiar enough with SE to know we had to chose answers. I just did. However at the time I was looking for a way to derive the QL/QC expresions from the resonance freq (i think?), I found it later myself so I didn't bother with the question. Just accepted it though, even if it was a partial answer. \$\endgroup\$
    – Yannick
    Jan 23, 2016 at 10:50

1 Answer 1


A few things first:

  1. I don't like your small signal models. They suck. (Professional opinion is that they're wrong because they don't state number 2)
  2. You need to know whether you're looking at high speed or low speed circuit when you use small signal models. Especially with mosfets as the capacitances in the model can look like open circuit or closed circuit depending on frequency.
  3. They move the current source impedance outside to a discrete resistor without an explanation... this is also another reason I don't like your small signal picture.

Okay. Griping about your text book over. Time to help you solve your problem. enter image description here

This is a much better picture. Note that this only works for small signals at low frequencies where the gate capacitance is essentially an open circuit. So, looking into the source we have a voltage dependent current source and ro. But goody for us, current sources are weird. You'd think that you should just throw out the current source because it has "infinite" impedance, but that'd be wrong here. Instead, it's transconductance gets inverted to become it's equivalent resistance. This is because we're looking into the arrow and not into the tail (you'll note that this only works with dependent current sources). So gmVgs current source becomes 1/gm resistance. Thus we now have two resistors, ro and (1/gm) in parallel with each other. So Ro = (1/gm)||ro. Yay, Magic!


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